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There are two charged spheres A and B having uniform charge density $+ ρ$ and $- ρ$ respectively. The intersection region has no charge. The radius of the sphere is R. The potential difference between the points O and O' is $V_{0} - V_{0'} = \frac{ρ R^{2}}{n ε_{0}}$ Points O and O’ are the centers of the two spheres. Then find n

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answer is 3.

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Detailed Solution

The potential at the points O and O' will be the super position of the potential due to both the charged spheres

$V_{O} = \frac{3 kQ}{2 R} - \frac{kQ}{R}$

$V_{O^{'}} = - \frac{3 kQ}{2 R} + \frac{kQ}{R}$

$V_{O} - V_{O^{'}} = \frac{kQ}{R} = \frac{1}{4 {πε}_{0}} \times \frac{\frac{4}{3} {πR}^{3} \times ρ}{R} = \frac{ρ R^{2}}{3 ε_{0}}$

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